Reynolds number Re = ρvL - μ — laminar vs turbulent criterion: Worked examples

2.2.19 · D3 · Physics › Fluid Mechanics › Reynolds number Re = ρvL - μ — laminar vs turbulent criterio

Shuru karne se pehle, un tools ki ek reminder jo hum poore page par use karte hain:

Scenario matrix

$R e$ ke baare mein har problem in cells mein se kisi ek mein aati hai. Neeche ke examples mein se har ek us cell ke saath tagged hai jise woh cover karta hai.

#	Scenario class	Kya cheez ise tricky banati hai	Example
A	Plain pipe — plug karo aur classify karo	SI units sahi karna	Ex 1
B	High-viscosity limit ( $μ$ huge)	$R e \to$ tiny, always laminar	Ex 2
C	High-speed / large-body limit	$R e \to$ huge, always turbulent	Ex 3
D	Solve for the threshold (find $v_{c}$ , or $L_{c}$ )	Rearranging the formula	Ex 4
E	Degenerate input: $v = 0$ or $L = 0$	$R e \to 0$ ; is "no flow" laminar?	Ex 5
F	Radius-vs-diameter trap	Which $L$ does the table assume?	Ex 6
G	Unit-conversion word problem (cm, mm, cP, air)	Convert before plugging	Ex 7
H	Kinematic-viscosity form ( $ν$ given, not $μ, ρ$ )	Use $R e = v L / ν$ directly	Ex 8
I	Exam twist: scaling / "what if we double it?"	Reason with ratios, no calculator	Ex 9
J	Two fluids, same $R e$ (dynamic similarity)	Match $R e$ across a model & real object	Ex 10
K	Right in the middle ( $2000 < R e < 4000$ )	Reading the transitional band	Ex 11

Woh regime rule jiske against hum classify karte hain (pipe convention, $L =$ diameter):

$R e$	Regime
$< 2000$	laminar
$2000$ – $4000$	transitional
$> 4000$	turbulent

Common mistake

Exact endpoints $R e = 2000$ aur $R e = 4000$ ko kaise padhein Kyun yeh ambiguous lagta hai: table teen bands dikhati hai, lekin $R e = 2000$ exactly kaun sa regime hai, ya $R e = 4000$ exactly? Kya $2000$ "laminar" hai ya "transitional"? Fix — yeh page jo convention use karta hai: do endpoints transitional band ke entrances hain, isliye hum unhe inclusively transition mein padhte hain: $R e \leq 2000 \Rightarrow laminar, 2000 < R e < 4000 \Rightarrow transitional, R e \geq 4000 \Rightarrow turbulent .$ Toh $R e = 2000$ laminar ka edge / transition ka start hai (Ex 4, Ex 6-radius), aur $R e = 4000$ turbulent ka edge / transition ka end hai (Ex 6-diameter). Kyunki $R e_{c}$ aise bhi fuzzy hai (parent note dekho), yeh boundary values ek side par argue karne ki bajaye "borderline" kehna sabse sahi hai — lekin exam consistency ke liye, upar boxed inequalities use karo.

Figure — Reynolds number Re = ρvL - μ — laminar vs turbulent criterion

Worked examples

Worked example Example 1 — Cell A: plain pipe, classify

Water ( $ρ = 1000 kg/m^{3}$ , $μ = 1.0 \times 1 0^{- 3} Pa\cdots$ ) $v = 2 m/s$ ki speed se $L = 0.05 m$ diameter ke pipe mein flow kar raha hai. Laminar hai ya turbulent?

Forecast: water, moderate speed, wide pipe — compute karne se pehle guess karo: turbulent hai ya laminar?

Formula likho: $R e = \frac{ρ v L}{μ}$ . Yeh step kyun? Humein number line par place karne ke liye ek single number chahiye; yahi woh quantity hai jo regime decide karti hai.
SI values substitute karo: $R e = \frac{1000 \times 2 \times 0.05}{1.0 \times 1 0 ^{- 3}}$ . Yeh step kyun? Sab kuch pehle se SI mein hai, toh units automatically cancel ho jaate hain aur hume ek pure number milta hai — koi unit bookkeeping ki zaroorat nahi.
Top simplify karo: $1000 \times 2 \times 0.05 = 100$ . Phir $R e = \frac{100}{1 0 ^{- 3}} = 100000$ . Yeh step kyun? Pehle numerator karo toh division ek clean "chhote number se divide karo = upar multiply karo" ban jaata hai.
Classify karo: $100000 ≫ 4000 \Rightarrow$ turbulent.

Verify karo: Units check — $\frac{( kg/m ^{3} ) ( m/s ) ( m )}{kg/(m\cdots)} = \frac{kg/(m\cdots)}{kg/(m\cdots)} = 1$ , dimensionless. ✓ Forecast (turbulent) match karta hai. Figure par yeh dot red band mein gehraai se baitha hai.

Worked example Example 2 — Cell B: high-viscosity limit

Glycerine ( $ρ = 1260 kg/m^{3}$ , $μ = 1.5 Pa\cdots$ ) $v = 2 m/s$ ki speed se same $L = 0.05 m$ pipe mein flow kar raha hai. Example 1 se compare karo.

Forecast: same speed, same pipe, lekin water se $1500 \times$ zyada viscous. Number line ka kaun sa cell?

$R e = \frac{1260 \times 2 \times 0.05}{1.5}$ . Yeh step kyun? Example 1 ke comparison mein sirf $ρ$ aur $μ$ change hue hain, toh hum expect karte hain ki answer $\frac{1260/1.5}{1000/1 0 ^{- 3}}$ ke factor se move karega — ek bahut bada shrink.
Numerator: $1260 \times 2 \times 0.05 = 126$ . Toh $R e = \frac{126}{1.5} = 84$ . Yeh step kyun? Bada $μ$ denominator mein hai; $0.001$ ki jagah $1.5$ se divide karna $R e$ ko $\approx 1500$ ke factor se crush kar deta hai.
Classify karo: $84 < 2000 \Rightarrow$ laminar — aslmein gehraai se laminar.

Verify karo: Ratio sanity: Ex 1 $1 0^{5}$ tha; $μ$ ko $1500$ se multiply karna aur $ρ$ ko $1.26$ se multiply karna predict karta hai $R e \approx 1 0^{5} \times \frac{1.26}{1500} \approx 84$ . ✓ "Viscosity peacemaker hai" rule confirm hota hai: woh bottom par rehti hai. Figure par yeh dot bahut left mein green band mein hai.

Worked example Example 3 — Cell C: high-speed / large-body limit

$L = 60 m$ lambai ki ek submarine $v = 10 m/s$ ki speed se seawater ( $ρ = 1025 kg/m^{3}$ , $μ = 1.1 \times 1 0^{- 3} Pa\cdots$ ) mein cruise kar rahi hai. $R e$ estimate karo.

Forecast: bahut bada body, theek-thaak speed, patla water — kya tum ek chhota ya astronomically bada number expect karte ho?

$R e = \frac{1025 \times 10 \times 60}{1.1 \times 1 0 ^{- 3}}$ . Yeh step kyun? Ek body ke liye (pipe nahi) $L$ body ki characteristic length hai; wahi formula phir bhi apply hota hai kyunki yeh ek hi do forces ka ratio hai.
Numerator: $1025 \times 10 \times 60 = 615000$ . Phir $R e = \frac{615000}{1.1 \times 1 0 ^{- 3}} \approx 5.59 \times 1 0^{8}$ . Yeh step kyun? Bada $L$ aur bada $v$ dono top ko multiply karte hain; tiny $μ$ divide karta hai — teen effects sab $R e$ ko upar push karte hain.
Classify karo: $5.6 \times 1 0^{8} ⋙ 4000 \Rightarrow$ strongly turbulent.

Verify karo: Order-of-magnitude: $\frac{1 0 ^{3} \cdot 10 \cdot 60}{1 0 ^{- 3}} = \frac{6 \times 1 0 ^{5}}{1 0 ^{- 3}} = 6 \times 1 0^{8}$ . ✓ Water/air mein koi bhi real vehicle $R e ≫ 1 0^{5}$ par rehta hai — rozana ki duniya turbulent hai.

Worked example Example 4 — Cell D: threshold speed ke liye solve karo

Example-1 ke water ke liye $L = 0.05 m$ pipe mein, kis speed $v_{c}$ par flow pehli baar transitional ( $R e = 2000$ ) banta hai?

Forecast: Example 1 mein $v = 2 m/s$ tha jisse $R e = 1 0^{5}$ mila. $R e = 2000$ tak aane ke liye humein $v$ chhota chahiye roughly $50 \times$ . $v_{c}$ guess karo.

$R e = \frac{ρ v L}{μ}$ ko $v$ ke liye rearrange karo: $v_{c} = \frac{R e μ}{ρ L}$ . Yeh step kyun? Unknown speed hai, toh hum $v$ ko isolate karte hain — dono sides ko $ρ L / μ$ se divide karo.
Substitute karo: $v_{c} = \frac{2000 \times 1 0 ^{- 3}}{1000 \times 0.05} = \frac{2}{50}$ . Yeh step kyun? Target $R e = 2000$ numerator mein daalo; baaki sab geometry/fluid hai jo hum fixed rakh rahe hain.
Compute karo: $v_{c} = 0.04 m/s$ . Exactly $R e = 2000$ par yeh laminar ka edge / transition ka start hai (upar endpoint box dekho).

Verify karo: Wapas plug karo: $R e = \frac{1000 \times 0.04 \times 0.05}{1 0 ^{- 3}} = \frac{2}{1 0 ^{- 3}} = 2000$ . ✓ Aur ratio check: $2 m/s$ ne $1 0^{5}$ diya; $2000$ , $1 0^{5}$ se $50 \times$ chhota hai, toh speed $50 \times$ chhoti honi chahiye: $2/50 = 0.04 m/s$ . ✓

Worked example Example 5 — Cell E: degenerate input,

$v \to 0$ Example-1 pipe mein pump off kar diya jaata hai, toh $v = 0$ . $R e$ kya hai, aur physically iska kya matlab hai?

Forecast: Kya "no flow" laminar hai, turbulent hai, ya kuch nahi?

$R e = \frac{ρ \cdot 0 \cdot L}{μ} = 0$ . Yeh step kyun? $v$ numerator mein ek factor hai; zero factor poore product ko zero kar deta hai — koi algebra tricks ki zaroorat nahi.
Interpret karo: $R e = 0 < 2000$ , toh yeh (trivially) laminar range mein hai. Yeh step kyun? Number line mein koi lower cutoff nahi hai; $2000$ se neeche ka har $R e$ laminar count hota hai, $0$ degenerate value bhi.
Physical caveat: $v = 0$ ke saath bilkul flow nahi hai — kuch bhi move nahi ho raha, toh "laminar" yahan sirf matlab hai "koi chaos nahi," jo sach hai par khali hai. Similarly $L = 0$ (pinched-shut pipe) $R e = 0$ deta hai: koi cross-section nahi, koi flow nahi. Yeh step kyun? Extreme inputs ko sirf arithmetic ke liye nahi, meaning ke liye bhi check karna chahiye — $R e = 0$ mathematically theek hai par ek static fluid describe karta hai.

Verify karo: Jab $v \to 0$ , $R e \to 0$ continuously; koi sudden jump nahi hai, toh "laminar at rest" reading self-consistent hai. ✓ Turbulence ko energy input chahiye; $v = 0$ ke saath koi nahi hai.

Worked example Example 6 — Cell F: radius-vs-diameter trap

Ek student ko bataya jaata hai ki pipe ki radius $r = 0.01 m$ hai, water $v = 0.2 m/s$ par ( $ρ = 1000$ , $μ = 1 0^{- 3}$ ). Woh $L = r$ plug karta hai aur ek answer paata hai; correct convention $L = D$ (diameter) use karti hai. Dono dikhao aur galat conclusion bhi.

Forecast: radius ki jagah diameter use karna $R e$ ko kis factor se change karta hai — aur kya yeh regime flip kar sakta hai?

Galat (radius): $R e_{wrong} = \frac{1000 \times 0.2 \times 0.01}{1 0 ^{- 3}} = \frac{2}{1 0 ^{- 3}} = 2000$ . Yeh step kyun? Exactly dikhao ki galat plug-in kya produce karta hai taaki hum error ki size dekh sakein.
Sahi (diameter $D = 2 r = 0.02 m$ ): $R e = \frac{1000 \times 0.2 \times 0.02}{1 0 ^{- 3}} = \frac{4}{1 0 ^{- 3}} = 4000$ . Yeh step kyun? Standard pipe table diameter ke liye calibrated hai; $D$ use karna $L$ ko double karta hai aur isliye $R e$ ko bhi double karta hai.
Consequence, endpoint box ke saath padho: radius plug $R e = 2000$ deta hai = laminar ka edge / transition ka start; correct diameter $R e = 4000$ deta hai = transition ka edge / turbulent ka start. Toh trap flow ko poore transitional band mein shift kar deta hai — "just leaving laminar" se "just entering turbulent" tak. Yeh step kyun? Isliye yeh mistake matter karti hai — yeh rounding error nahi hai; dono endpoints borderline hain, lekin woh transition zone ke opposite ends par baithe hain, toh practical verdict change ho jaata hai.

Verify karo: $R e_{D} / R e_{r} = D / r = 2$ exactly, kyunki $R e \propto L$ . ✓ Apni table trust karne se pehle hamesha confirm karo ki woh kaun sa $L$ assume karti hai.

Worked example Example 7 — Cell G: unit-conversion word problem (air, cP defined ke saath)

Air $50 cm$ diameter ke ventilation duct mein $v = 3 m/s$ se blow kar rahi hai. Air ka $ρ = 1.2 kg/m^{3}$ hai aur viscosity $μ = 18 cP$ quote ki gayi hai. Classify karo.

cP kya hai? Centipoise (cP) ek common non-SI viscosity unit hai. Conversion hai $1 cP = 1 0^{- 3} Pa\cdots$ (handy baat yeh hai ki room temperature par pure water $\approx 1 cP$ hai). Toh $18 cP = 18 \times 1 0^{- 3} cP \dots$ — nahi: $18 cP = 18 \times 1 0^{- 3} Pa\cdots$ ? Dhyan se: $18 cP = 18 \times 1 0^{- 3} Pa\cdots = 0.018 Pa\cdots$ air ke liye bahut thick hoga. Air $\approx 0.018 cP$ hai, yaani $μ_{air} = 0.018 cP = 1.8 \times 1 0^{- 5} Pa\cdots$ . Hum neeche is correct value ka use karte hain.

Forecast: air light aur thin hai — kya light $ρ$ (chhota numerator) ya thin $μ$ (chhota denominator) jeetta hai?

Viscosity ko SI mein convert karo: $μ = 0.018 cP \times \frac{1 0 ^{- 3} Pa\cdots}{1 cP} = 1.8 \times 1 0^{- 5} Pa\cdots$ . Yeh step kyun? Formula ko Pa·s chahiye; $μ$ ko cP mein chhod dena $R e$ ko $1 0^{3}$ se galat kar dega. Plug karne se pehle har non-SI quantity convert karo.
Length convert karo: $50 cm = 0.50 m$ . Yeh step kyun? cm aur m ko mix karna $R e$ ko $100$ se multiply kar dega aur answer kharab kar dega.
Substitute karo: $R e = \frac{1.2 \times 3 \times 0.50}{1.8 \times 1 0 ^{- 5}}$ . Yeh step kyun? Ab har quantity SI mein hai, toh units cancel ho jaate hain aur hume pure number milta hai.
Numerator: $1.2 \times 3 \times 0.50 = 1.8$ . Phir $R e = \frac{1.8}{1.8 \times 1 0 ^{- 5}} = 100000$ . Yeh step kyun? Tiny denominator ( $1 0^{- 5}$ ) chhote numerator par decisively jeet jaata hai — air ki thinness dominate karti hai.
Classify karo: $100000 > 4000 \Rightarrow$ turbulent. Real ventilation ducts wakai turbulent hote hain (isliye woh air acchi tarah mix karte hain).

Verify karo: Agar hum convert karna bhool jaate aur $L = 50$ (cm mein) use karte, toh $R e = 1 0^{7}$ milta — clearly absurd ek duct ke liye, ek red flag ki conversion skip ki gayi. Correct value $1 0^{5}$ . ✓

Worked example Example 8 — Cell H: kinematic-viscosity form

Oil ek pipe mein $L = 0.03 m$ ke saath $v = 0.5 m/s$ par flow kar raha hai. Tumhe sirf uski kinematic viscosity $ν = 1.0 \times 1 0^{- 4} m^{2} / s$ di gayi hai (density aur $μ$ nahi di gayi). $R e$ dhundho.

Forecast: $ν$ pehle se $μ / ρ$ ko bundle karta hai — kya tumhe $ρ$ ki zaroorat bhi hai?

Doosri form use karo: $R e = \frac{v L}{ν}$ . Yeh step kyun? Jab $ν$ directly supply ki jaati hai, yeh form $ρ$ aur $μ$ ko bilkul skip kar deta hai — kam numbers, kam errors.
Substitute karo: $R e = \frac{0.5 \times 0.03}{1.0 \times 1 0 ^{- 4}} = \frac{0.015}{1 0 ^{- 4}}$ . Yeh step kyun? Numerator $v L = 0.015 m^{2} / s$ compute karo; $ν$ (bhi $m^{2} / s$ ) se divide karna units ko pure number mein cancel kar deta hai.
Compute karo: $R e = 150$ . Yeh step kyun? $0.015$ ko $0.0001$ se divide karo = $1 0^{4}$ se multiply karo phir $1.5$ se → $150$ .
Classify karo: $150 < 2000 \Rightarrow$ laminar.

Verify karo: Dono forms agree karte hain kyunki $ν = μ / ρ$ , toh $\frac{v L}{ν} = \frac{ρ v L}{μ}$ . Units: $\frac{( m/s ) ( m )}{m ^{2} / s} = 1$ . ✓

Worked example Example 9 — Cell I: scaling twist (calculator nahi)

Ek flow $R e = 1800$ par laminar hai. Engineers pipe diameter double karte hain aur flow speed triple karte hain, same fluid rakh ke. Naya $R e$ ? Naya regime?

Forecast: koi arithmetic karne se pehle $R e$ par multiplier guess karo.

Note karo $R e \propto v L$ ( $ρ, μ$ fixed ke saath). Doubling $L$ aur tripling $v$ $R e$ ko $2 \times 3 = 6$ se multiply karta hai. Yeh step kyun? Kyunki $R e$ $v$ aur $L$ dono mein linear hai, changes sirf multiply hote hain — actual $ρ, μ$ values ki zaroorat nahi.
Naya $R e = 6 \times 1800 = 10800$ . Yeh step kyun? Known starting value par multiplier apply karo.
Classify karo: $10800 > 4000 \Rightarrow$ turbulent. Flow laminar se seedha transition ke through turbulent mein flip ho gaya. Yeh step kyun? Regime change report karo — twist ka point yahi hai.

Verify karo: Linearity check: agar hum dono ko halve karte, factor $\frac{1}{4}$ , giving $450$ — phir bhi laminar, consistent with " $R e$ scales as $v L$ ." ✓

Worked example Example 10 — Cell J: dynamic similarity (do fluids, same

$R e$ ) Ek model car ( $L_{m} = 0.4 m$ ) ko ek water tunnel ( $ν_{water} = 1.0 \times 1 0^{- 6} m^{2} / s$ ) mein test kiya jaata hai taaki ek real car ( $L_{r} = 4 m$ ) jo air ( $ν_{air} = 1.5 \times 1 0^{- 5} m^{2} / s$ ) mein $v_{r} = 30 m/s$ se chalti hai usse mimic kar sake. Kaun si test speed $v_{m}$ dono flows ko match karaati hai ( $R e_{m} = R e_{r}$ )?

Forecast: water kinematically $15 \times$ kam viscous hai lekin model $10 \times$ chhota hai — kya yeh effects ek fast ya slow test speed tak add up hote hain?

Real-car Reynolds number: $R e_{r} = \frac{v _{r} L _{r}}{ν _{air}} = \frac{30 \times 4}{1.5 \times 1 0 ^{- 5}} = \frac{120}{1.5 \times 1 0 ^{- 5}} = 8 \times 1 0^{6}$ . Yeh step kyun? Dynamic similarity ka matlab hai ki dimensionless number equal hona chahiye; pehle target $R e$ dhundho jise match karna hai.
$R e_{m} = R e_{r}$ set karo aur model speed ke liye solve karo: $v_{m} = \frac{R e _{r} ν _{water}}{L _{m}} = \frac{8 \times 1 0 ^{6} \times 1.0 \times 1 0 ^{- 6}}{0.4}$ . Yeh step kyun? $R e_{m} = v_{m} L_{m} / ν_{water}$ rearrange karna sirf unknown $v_{m}$ ko isolate karta hai.
Compute karo: numerator $8 \times 1 0^{6} \times 1 0^{- 6} = 8$ ; phir $v_{m} = \frac{8}{0.4} = 20 m/s$ . Yeh step kyun? Yeh test speed model ko real car ke same $R e$ deti hai, toh flow pattern physically identical hai.
Classify karo: dono flows $R e = 8 \times 1 0^{6} > 4000$ share karte hain $\Rightarrow$ model aur real car dono ke liye turbulent — yehi poora point hai: woh same regime mein hain. Yeh step kyun? Equal $R e$ ⇒ same regime, same wake, same drag coefficient.

Verify karo: Dono sides check karo: $R e_{m} = \frac{20 \times 0.4}{1 0 ^{- 6}} = \frac{8}{1 0 ^{- 6}} = 8 \times 1 0^{6} = R e_{r}$ . ✓ Same $R e$ ⇒ same regime — exactly isliye wind-tunnel testing kaam karta hai. Dekho Drag force and drag coefficient.

Worked example Example 11 — Cell K: bilkul beech mein (transitional band)

Water ( $ρ = 1000 kg/m^{3}$ , $μ = 1.0 \times 1 0^{- 3} Pa\cdots$ ) $v = 0.15 m/s$ ki speed se $L = 0.02 m$ diameter ke pipe mein flow kar raha hai. Kaun sa regime?

Forecast: slow-ish water narrow-ish pipe mein — kya yeh cleanly laminar mein, cleanly turbulent mein, ya awkwardly beech mein land karega?

$R e = \frac{1000 \times 0.15 \times 0.02}{1.0 \times 1 0 ^{- 3}}$ . Yeh step kyun? Same plug-and-classify Ex 1 ki tarah, lekin choose kiya gaya hai taaki answer do thresholds ke beech pade — woh case jo table ki middle row describe karti hai.
Numerator: $1000 \times 0.15 \times 0.02 = 3$ . Phir $R e = \frac{3}{1 0 ^{- 3}} = 3000$ . Yeh step kyun? Pehle top compute karo, phir small $μ$ se divide karo.
Endpoint box ke saath classify karo: $2000 < 3000 < 4000 \Rightarrow$ transitional. Yeh step kyun? $3000$ strictly yellow band ke andar hai — na $\leq 2000$ na $\geq 4000$ — toh yeh genuinely middle regime mein hai.
"Transitional" ka physical matlab: flow flicker karta hai — smooth laminar motion ke patches appear hote hain aur intermittently turbulent bursts mein toot jaate hain, aur exact behaviour pipe roughness aur vibration ke liye sensitive hai. Engineers is band mein design karne se bachte hain kyunki yeh unpredictable hai. Yeh step kyun? Regime ka naam lena kaafi nahi; transitional band intermittency se define hota hai, aur woh exam-relevant physical fact hai.

Verify karo: Wapas plug karo aur figure par locate karo: $R e = 3000$ yellow band mein squarely baitha hai, $2000$ aur $4000$ ticks ke beech. Ise laminar push karne ke liye tumhe $v < 0.1 m/s$ chahiye ( $R e < 2000$ ); turbulent push karne ke liye, $v > 0.2 m/s$ ( $R e > 4000$ ). ✓

Recall Saare cells mein quick self-test

Sirf speed double karna $R e$ ke saath kya karta hai? ::: Ise double karta hai ( $v$ mein linear). $L$ ke roop mein radius ki jagah diameter switch karna kya karta hai? ::: $R e$ double ho jaata hai. $v = 0$ $R e =$ ? deta hai aur kaun sa regime? ::: $R e = 0$ , trivially laminar (lekin koi flow nahi). $μ, ρ$ ki jagah $ν$ diya gaya ho toh kaun sa formula? ::: $R e = v L / ν$ . Equal $R e$ wale do flows kya kahlaate hain? ::: Dynamically similar (same regime & pattern). $R e = 3000$ kaun sa regime hai? ::: Transitional (2000 aur 4000 ke beech). Exact endpoint $R e = 2000$ ko kaise padho? ::: Borderline — laminar ka edge / transition ka start.

Connections

Parent topic — woh derivation jise yeh examples apply karte hain.
Viscosity and Newton's law of viscosity — jahan se $μ$ aur $ν = μ / ρ$ aate hain.
Poiseuille's law — sirf laminar law (Ex 2, 5, 8 ke liye valid).
Stokes' law and terminal velocity — low- $R e$ sphere case (Ex 2 ke relatives).
Bernoulli's principle — $R e \to \infty$ inviscid idealisation (Ex 3).
Dimensional analysis — kyun $R e$ match karna (Ex 10) similarity guarantee karta hai.
Drag force and drag coefficient — $C_{d} (R e)$ , Ex 10 ki similarity ka payoff.